Calculating Elevator Motor Horsepower

This problem asks us to find the minimum horsepower required for a motor to lift an elevator at a constant velocity. To lift the elevator, the motor needs to apply an upward force that is at least equal to the weight of the elevator. Since the elevator is being lifted at a constant velocity, the net force on it must be zero. This means the upward force applied by the motor must exactly balance the downward gravitational force (the weight).

Understanding the Physics

• Constant Velocity: When an object moves at a constant velocity, its acceleration is zero. According to Newton's second law ($\text{F}_{net} = \text{m} \times \text{a}$), if acceleration is zero, the net force acting on the object must also be zero.
• Forces on the Elevator: There are two main forces acting on the elevator:
  • Downward force due to gravity (Weight).
  • Upward force applied by the motor.
• Balancing Forces: For zero net force (constant velocity), the upward force must equal the downward force. Thus, the force the motor must exert is equal to the weight of the elevator.
• Power: Power is the rate at which work is done, or the rate at which energy is transferred. For an object moving at a constant velocity ($\text{v}$) under the action of a constant force ($\text{F}$) in the direction of motion, power ($\text{P}$) is given by the formula $\text{P} = \text{F} \times \text{v}$.

Step-by-Step Calculation of Motor Power

First, we need to calculate the weight of the elevator, which is the force the motor must overcome.

Given:

• Mass of the elevator ($\text{m}$) = 400 kg
• Acceleration due to gravity ($\text{g}$) = 9.8 m/s$^2$
• Constant velocity ($\text{v}$) = 0.20 m/s

Step 1: Calculate the weight of the elevator (Force).

Weight ($\text{F}$) = Mass ($\text{m}$) $\times$ acceleration due to gravity ($\text{g}$)

$\text{F} = 400 \text{ kg} \times 9.8 \text{ m/s}^2$

$\text{F} = 3920 \text{ N}$

The motor must exert an upward force of 3920 N to lift the elevator at a constant velocity.

Step 2: Calculate the power required in Watts.

Power ($\text{P}$) = Force ($\text{F}$) $\times$ velocity ($\text{v}$)

$\text{P} = 3920 \text{ N} \times 0.20 \text{ m/s}$

$\text{P} = 784 \text{ Watts}$

Step 3: Convert power from Watts to Horsepower.

We need to convert the power calculated in Watts to horsepower (hp). The conversion factor is approximately 1 horsepower = 746 Watts.

Power in hp = $\frac{\text{Power in Watts}}{746 \text{ Watts/hp}}$

Power in hp = $\frac{784}{746}$

Power in hp $\approx 1.0509$ hp

The minimum horsepower required for the motor is approximately 1.05 hp.

Comparing with Options

Let's look at the given options:

• 2.0 hp
• 0.5 hp
• 1.05 hp
• 5.0 hp

Our calculated value, approximately 1.05 hp, matches option 3.

Summary of Calculation

The minimum horsepower of the motor to be used is approximately 1.05 hp.

Revision Table: Key Concepts

Additional Information: Power and Work

Power is a fundamental concept in physics that tells us how quickly work is being done. Work is done when a force causes displacement. Lifting the elevator requires the motor to do work against gravity. The faster the elevator is lifted, the more power is required, because the same amount of work is done in less time.

• Work (W): Defined as $\text{W} = \text{F} \times \text{d} \times \cos(\theta)$, where $\text{F}$ is the force, $\text{d}$ is the displacement, and $\theta$ is the angle between the force and displacement vectors. When lifting, the force and displacement are in the same direction ($\theta = 0^\circ$, $\cos(0^\circ) = 1$), so $\text{W} = \text{F} \times \text{d}$.
• Power (P): Defined as the rate of doing work, $\text{P} = \frac{\text{W}}{\text{t}}$. If work is done over a time $\text{t}$, the average power is Work divided by time. For constant force and velocity, as in this elevator problem, we can use the relation $\text{P} = \text{F} \times \text{v}$. This is derived from $\text{P} = \frac{\text{W}}{\text{t}} = \frac{\text{F} \times \text{d}}{\text{t}}$. Since velocity $\text{v} = \frac{\text{d}}{\text{t}}$, we get $\text{P} = \text{F} \times \text{v}$.
• Minimum Horsepower: The question asks for the minimum horsepower. This corresponds to the power needed to lift the elevator at the specified constant velocity without considering any energy losses due to friction. If friction were present, the motor would need to provide additional force (and thus more power) to overcome friction as well.

This problem is a direct application of the power formula $\text{P} = \text{F} \times \text{v}$ in the context of lifting an object against gravity.